3
$\begingroup$

I want to find a semi-constructive example of a unitary commutative ring without any maximal ideals assuming that axiom of choice is incorrect and/or a model of $\sf ZF$ where we have such a concrete ring.

This question is similar to the questions Vector space bases without axiom of choice and A confusion about Axiom of Choice and existence of maximal ideals..

What I tried is to use $\mathbb{R}$ as a $\mathbb{Q}$ vector space without a basis and try to construct some chains of ideals on a related ring and try to show that a maximal ideal corresponds to a basis but didn't achieve much.

Thank you in advance!

|cite|improve this question
$\endgroup$
  • $\begingroup$ As far as I know, existence of maximal ideals in rings is properly weaker than choice. $\endgroup$ – egreg Jul 7 '19 at 13:59
  • 1
    $\begingroup$ @egreg math.stackexchange.com/questions/317028/… - they seem equal $\endgroup$ – Konformist Liberal Jul 7 '19 at 14:24
  • 1
    $\begingroup$ I stand corrected. $\endgroup$ – egreg Jul 7 '19 at 14:28
  • 1
    $\begingroup$ Well. You can either construct special models with particular examples, and that normally involves techniques like forcing and stuff; or you can sort of repeat the proof that the existence of maximal ideals implies choice and start with a set that cannot be well-ordered as a counterexample. This set can be $\Bbb R$ in models where there is no Hamel basis, for example. $\endgroup$ – Asaf Karagila♦ Jul 7 '19 at 14:46
  • 1
    $\begingroup$ Near-duplicate: math.stackexchange.com/questions/2182014/… $\endgroup$ – Eric Wofsey Jul 7 '19 at 18:29
2
$\begingroup$

Let $k$ be a field, let $I\subset k^{\mathbb{N}}$ be the ideal of sequences that are eventually zero, and let $S=k^{\mathbb{N}}/I$. Then maximal ideals in $S$ are in bijection with nonprincipal ultrafilters on $\mathbb{N}$. In particular, in any model of ZF in which there are no nonprincipal ultrafilters on $\mathbb{N}$, there will be no maximal ideals in $S$.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.